Q53 / 150 · MHT CET 2012 · Mathematics

If the four positive integers are selected randomly from the set of positive integers, then the probability that the number $1,3,7$ and 9 are in the unit place in the product of 4 -digit, so selected is

  • A
    $\frac{7}{625}$
  • B
    $\frac{2}{5}$
  • C
    $\frac{5}{625}$
  • Correct answer
    D
    $\frac{16}{625}$

Solution

The number of digits on unit place of any number $=10$ $\therefore$ $$ n(S)=10 $$ The necessary condition for becoming the digits $1,3,5$ or 7 at the unit place of product of four numbers that the digits $1,3,5$ or 7 at unit place of every number. $$ \begin{array}{ll} \therefore & n(A)=4 \\ \therefore & P(A)=\frac{4}{10}=\frac{2}{5} \end{array} $$ So, required $y=\left(\frac{2}{5}\right)^{4}=\frac{16}{625}$

Hence, the correct answer is option: 'D'.

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